Problem 19 - Calculus: Early Transcendentals

Question

Find a power series representation for the function and determine the radius of convergence. \( f(x) = \frac {1 + x}{(1 - x)^2} \)

Step-by-Step Solution

Answer

The power series is \( \sum_{n=0}^{\infty} (2n+1)x^n \) with a radius of convergence 1.

Step 1: Identify the Basic Geometric Series

The geometric series is given by \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \) for \(|x| < 1\). Our goal is to manipulate \(\frac{1+x}{(1-x)^2}\) into a form where we can use this series as a component.

Step 2: Derivative of the Basic Series

Differentiate the geometric series \( \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n \) with respect to \( x \) to find \( \frac{1}{(1-x)^2} \). This yields \( \frac{d}{dx}\left( \frac{1}{1-x} \right) = \sum_{n=1}^{\infty} nx^{n-1} = \frac{1}{(1-x)^2} \). Therefore, \( \frac{1}{(1-x)^2} = \sum_{n=0}^{\infty} (n+1)x^n \).

Step 3: Multiply by \(1 + x\)

Multiply the result from Step 2 by \(1 + x\) to determine \( f(x) \):\[(1 + x) \cdot \frac{1}{(1-x)^2} = (1+x) \cdot \sum_{n=0}^{\infty} (n+1)x^n = \sum_{n=0}^{\infty} (n+1)x^n + \sum_{n=0}^{\infty} (n+1)x^{n+1}.\]

Step 4: Combine the Series

Combine the two series to form a single series:For the term \( \sum_{n=0}^{\infty} (n+1)x^n + \sum_{n=0}^{\infty} (n+1)x^{n+1} \), change the index on the second series to match the first series:\[\sum_{n=0}^{\infty} (n+1)x^n + \sum_{n=1}^{\infty} nx^n = (1) + \sum_{n=1}^{\infty} ((n+1) + n)x^n.\] Evaluate to obtain: \[sum_{n=0}^{\infty} (2n+1)x^n.\]

Step 5: Determine the Radius of Convergence

Since the series are based on the geometric series with a radius of convergence \(|x| < 1\), all transformations (differentiation and multiplication by \(1+x\)) do not affect this interval. Therefore, the radius of convergence is 1.

Key Concepts

Radius of Convergence

When working with power series, determining the radius of convergence is crucial. It tells us the range of values for which the series converges to a function. For a geometric series like \[\sum_{n=0}^{\infty} x^n\]in \(\frac{1}{1-x}\), it converges when \(|x| < 1\).
This means the radius of convergence is 1.
To find the radius of convergence for more complex series, we can often rely on manipulation. Operations like differentiation or multiplication do not typically affect this radius. Therefore, even after transforming the geometric series into \(\frac{1+x}{(1-x)^2}\), the radius remains 1. This idea offers consistency and simplifies handling complex power series.

Geometric Series

The geometric series is a fundamental building block in calculus.It's described by\[\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\]for\(|x| < 1\). This simple form makes it easy to use for further transformations and operations.
Understanding the behavior of a geometric series helps when manipulating functions into power series form.In particular, noticing patterns in exponents and coefficients allows students to adjust the series to fit more complex functions.
The power of the geometric series lies in its simplicity and flexibility, providing a versatile tool for calculus problems. It's often the first step when finding power series representations for more complicated functions.

Differentiation

Differentiation in the context of power series involves modifying the series to obtain new expressions. When differentiating the geometric series \(\frac{1}{1-x}\), we find\[\frac{d}{dx}\left( \sum_{n=0}^{\infty} x^n \right) = \sum_{n=1}^{\infty} n x^{n-1}\].This results in \(\frac{1}{(1-x)^2}\).
Differentiating is key to tackling \(\frac{1+x}{(1-x)^2}\), as it helps us transform the initial series into a more complex form. Repeated derivatives can further modify the function, giving a deeper insight into its behavior and convergence.
Differentiation also preserves the radius of convergence, making it a safe operation when analyzing the range of convergence for a power series.

Series Manipulation

Series manipulation involves using algebraic operations to transform or combine series into a desired form. For instance, when expressing \(f(x)=\frac{1+x}{(1-x)^2}\) as a power series, we first manipulate the known series for \(\frac{1}{(1-x)^2}\).
By multiplying the series \(\sum_{n=0}^{\infty} (n+1)x^n\) by \(1 + x\), we engage in a series of algebraic steps. These include adjusting indices and combining sums, ultimately leading to a unified series.
Thorough understanding of series manipulation simplifies complex functions, enabling solutions through methodical steps. This concept forms the backbone of power series conversion, providing an elegant strategy to arrive at solutions.